问题描述
我的程序最近在运行时遇到奇怪的段错误. 我想知道以前是否有人遇到过此错误，以及如何解决该错误.这是更多信息:
My program recently encountered a weird segfault when running. I want to know if somebody had met this error before and how it could be fixed. Here is more info:
基本信息:
- CentOS 5.2，内核版本为2.6.18
- g ++(GCC)4.1.2 20080704(Red Hat 4.1.2-50)
- CPU:英特尔x86家族
- libstdc ++.so.6.0.8
- 我的程序将启动多个线程来处理数据.段错误发生在其中一个线程中.
- 尽管它是一个多线程程序，但段错误似乎发生在本地std :: string对象上.稍后，我将在代码片段中对此进行显示.
- 该程序是使用-g，-Wall和-fPIC编译的，并且没有-O2或其他优化选项.
核心转储信息:
Core was generated by `./myprog'.
Program terminated with signal 11, Segmentation fault.
#0 0x06f6d919 in __gnu_cxx::__exchange_and_add(int volatile*, int) () from /usr/lib/libstdc++.so.6
(gdb) bt
#0 0x06f6d919 in __gnu_cxx::__exchange_and_add(int volatile*, int) () from /usr/lib/libstdc++.so.6
#1 0x06f507c3 in std::basic_string<char, std::char_traits<char>, std::allocator<char> >::assign(std::basic_string<char, std::char_traits<char>, std::allocator<char> > const&) () from /usr/lib/libstdc++.so.6
#2 0x06f50834 in std::basic_string<char, std::char_traits<char>, std::allocator<char> >::operator=(std::basic_string<char, std::char_traits<char>, std::allocator<char> > const&) () from /usr/lib/libstdc++.so.6
#3 0x081402fc in Q_gdw::ProcessData (this=0xb2f79f60) at ../../../myprog/src/Q_gdw/Q_gdw.cpp:798
#4 0x08117d3a in DataParser::Parse (this=0x8222720) at ../../../myprog/src/DataParser.cpp:367
#5 0x08119160 in DataParser::run (this=0x8222720) at ../../../myprog/src/DataParser.cpp:338
#6 0x080852ed in Utility::__dispatch (arg=0x8222720) at ../../../common/thread/Thread.cpp:603
#7 0x0052c832 in start_thread () from /lib/libpthread.so.0
#8 0x00ca845e in clone () from /lib/libc.so.6
请注意，段故障始于 basic_string :: operator =().
相关代码:(我展示的代码超出了可能需要的数量，请暂时忽略编码风格.)
The related code:(I've shown more code than that might be needed, and please ignore the coding style things for now.)
int Q_gdw::ProcessData()
{
char tmpTime[10+1] = {0};
char A01Time[12+1] = {0};
std::string tmpTimeStamp;
// Get the timestamp from TP
if((m_BackFrameBuff[11] & 0x80) >> 7)
{
for (i = 0; i < 12; i++)
{
A01Time[i] = (char)A15Result[i];
}
tmpTimeStamp = FormatTimeStamp(A01Time, 12); // Segfault occurs on this line
这是此FormatTimeStamp方法的原型:
And here is the prototype of this FormatTimeStamp method:
std::string FormatTimeStamp(const char *time, int len)
我认为这样的字符串分配操作应该是一种常用的操作，但是我只是不明白为什么这里可能发生段错误.
I think such string assignment operations should be a kind of commonly used one, but I just don't understand why a segfault could occurr here.
我调查的内容:
我已经在网上搜索了答案.我在此处进行了查看.答复说，尝试使用定义的_GLIBCXX_FULLY_DYNAMIC_STRING宏重新编译程序.我尝试过，但仍然发生崩溃.
I've searched on the web for answers. I looked at here. The reply says try to recompile the program with _GLIBCXX_FULLY_DYNAMIC_STRING macro defined. I tried but the crash still happens.
我还此处.它还说要使用_GLIBCXX_FULLY_DYNAMIC_STRING重新编译该程序，但是作者似乎正在处理我的另一个问题，因此我认为他的解决方案不适合我.
I also looked at here. It also says to recompile the program with _GLIBCXX_FULLY_DYNAMIC_STRING, but the author seems to be dealing with a different problem with mine, thus I don't think his solution works for me.
已于2011年8月15日更新
大家好，这是FormatTimeStamp的原始代码.我了解编码看起来不太好(例如，太多的魔术数字.)，但让我们首先关注崩溃问题.
Hi guys, here is the original code of this FormatTimeStamp. I understand the coding doesn't look very nice(too many magic numbers, for instance..), but let's focus on the crash issue first.
string Q_gdw::FormatTimeStamp(const char *time, int len)
{
string timeStamp;
string tmpstring;
if (time) // It is guaranteed that "time" is correctly zero-terminated, so don't worry about any overflow here.
tmpstring = time;
// Get the current time point.
int year, month, day, hour, minute, second;
#ifndef _WIN32
struct timeval timeVal;
struct tm *p;
gettimeofday(&timeVal, NULL);
p = localtime(&(timeVal.tv_sec));
year = p->tm_year + 1900;
month = p->tm_mon + 1;
day = p->tm_mday;
hour = p->tm_hour;
minute = p->tm_min;
second = p->tm_sec;
#else
SYSTEMTIME sys;
GetLocalTime(&sys);
year = sys.wYear;
month = sys.wMonth;
day = sys.wDay;
hour = sys.wHour;
minute = sys.wMinute;
second = sys.wSecond;
#endif
if (0 == len)
{
// The "time" doesn't specify any time so we just use the current time
char tmpTime[30];
memset(tmpTime, 0, 30);
sprintf(tmpTime, "%d-%d-%d %d:%d:%d.000", year, month, day, hour, minute, second);
timeStamp = tmpTime;
}
else if (6 == len)
{
// The "time" specifies "day-month-year" with each being 2-digit.
// For example: "150811" means "August 15th, 2011".
timeStamp = "20";
timeStamp = timeStamp + tmpstring.substr(4, 2) + "-" + tmpstring.substr(2, 2) + "-" +
tmpstring.substr(0, 2);
}
else if (8 == len)
{
// The "time" specifies "minute-hour-day-month" with each being 2-digit.
// For example: "51151508" means "August 15th, 15:51".
// As the year is not specified, the current year will be used.
string strYear;
stringstream sstream;
sstream << year;
sstream >> strYear;
sstream.clear();
timeStamp = strYear + "-" + tmpstring.substr(6, 2) + "-" + tmpstring.substr(4, 2) + " " +
tmpstring.substr(2, 2) + ":" + tmpstring.substr(0, 2) + ":00.000";
}
else if (10 == len)
{
// The "time" specifies "minute-hour-day-month-year" with each being 2-digit.
// For example: "5115150811" means "August 15th, 2011, 15:51".
timeStamp = "20";
timeStamp = timeStamp + tmpstring.substr(8, 2) + "-" + tmpstring.substr(6, 2) + "-" + tmpstring.substr(4, 2) + " " +
tmpstring.substr(2, 2) + ":" + tmpstring.substr(0, 2) + ":00.000";
}
else if (12 == len)
{
// The "time" specifies "second-minute-hour-day-month-year" with each being 2-digit.
// For example: "305115150811" means "August 15th, 2011, 15:51:30".
timeStamp = "20";
timeStamp = timeStamp + tmpstring.substr(10, 2) + "-" + tmpstring.substr(8, 2) + "-" + tmpstring.substr(6, 2) + " " +
tmpstring.substr(4, 2) + ":" + tmpstring.substr(2, 2) + ":" + tmpstring.substr(0, 2) + ".000";
}
return timeStamp;
}
2011年8月19日更新
此问题终于得到解决.实际上，FormatTimeStamp()函数与根本原因无关. segfault是由本地char缓冲区的写溢出引起的.
This problem has finally been addressed and fixed. The FormatTimeStamp() function has nothing to do with the root cause, in fact. The segfault is caused by a writing overflow of a local char buffer.
此问题可以通过以下更简单的程序重现(请暂时忽略某些变量的错误命名):
This problem can be reproduced with the following simpler program(please ignore the bad namings of some variables for now):
(与"g ++ -Wall -g main.cpp一起编译")
(Compiled with "g++ -Wall -g main.cpp")
#include <string>
#include <iostream>
void overflow_it(char * A15, char * A15Result)
{
int m;
int t = 0,i = 0;
char temp[3];
for (m = 0; m < 6; m++)
{
t = ((*A15 & 0xf0) >> 4) *10 ;
t += *A15 & 0x0f;
A15 ++;
std::cout << "m = " << m << "; t = " << t << "; i = " << i << std::endl;
memset(temp, 0, sizeof(temp));
sprintf((char *)temp, "%02d", t); // The buggy code: temp is not big enough when t is a 3-digit integer.
A15Result[i++] = temp[0];
A15Result[i++] = temp[1];
}
}
int main(int argc, char * argv[])
{
std::string str;
{
char tpTime[6] = {0};
char A15Result[12] = {0};
// Initialize tpTime
for(int i = 0; i < 6; i++)
tpTime[i] = char(154); // 154 would result in a 3-digit t in overflow_it().
overflow_it(tpTime, A15Result);
str.assign(A15Result);
}
std::cout << "str says: " << str << std::endl;
return 0;
}
以下是我们应该记住的两个事实:1).我的机器是Intel x86机器，因此使用的是Little Endian规则.因此，对于int类型的变量"m"(其值为10)，其内存布局可能如下所示:
Here are two facts we should remember before going on:1). My machine is an Intel x86 machine so it's using the Little Endian rule. Therefore for a variable "m" of int type, whose value is, say, 10, it's memory layout might be like this:
Starting addr：0xbf89bebc: m(byte#1): 10
0xbf89bebd: m(byte#2): 0
0xbf89bebe: m(byte#3): 0
0xbf89bebf: m(byte#4): 0
2).上面的程序在主线程中运行.当涉及到overflow_it()函数时，线程堆栈中的变量布局如下所示(仅显示重要变量):
2). The program above runs within the main thread. When it comes to the overflow_it() function, the variables layout in the thread stack looks like this(which only shows the important variables):
0xbfc609e9 : temp[0]
0xbfc609ea : temp[1]
0xbfc609eb : temp[2]
0xbfc609ec : m(byte#1) <-- Note that m follows temp immediately. m(byte#1) happens to be the byte temp[3].
0xbfc609ed : m(byte#2)
0xbfc609ee : m(byte#3)
0xbfc609ef : m(byte#4)
0xbfc609f0 : t
...(3 bytes)
0xbfc609f4 : i
...(3 bytes)
...(etc. etc. etc...)
0xbfc60a26 : A15Result <-- Data would be written to this buffer in overflow_it()
...(11 bytes)
0xbfc60a32 : tpTime
...(5 bytes)
0xbfc60a38 : str <-- Note the str takes up 4 bytes. Its starting address is **16 bytes** behind A15Result.
我的分析:
1). m是overflow_it()中的一个计数器，其值在每个for循环中均增加1，并且其最大值假定不大于6.因此，它的值可以完全存储在m(byte#1)(记住它是Little Endian)中恰好是温度 3 .
1). m is a counter in overflow_it() whose value is incremented by 1 at each for loop and whose max value is supposed not greater than 6. Thus it's value could be stored completely in m(byte#1)(remember it's Little Endian) which happens to be temp3.
2).在越野车行中:当t是3位整数(例如109)时，sprintf()调用将导致缓冲区溢出，因为将数字109序列化为字符串"109"实际上需要4个字节:'1' ，"0"，"9"和终止符"\ 0".因为temp []仅分配了3个字节，所以最终的'\ 0'肯定会写入temp 3 ，它只是m(byte#1)，不幸的是存储了m的值.结果，每次将m的值重置为0.
2). In the buggy line: When t is a 3-digit integer, such as 109, then the sprintf() call would result in a buffer overflow, because serializing the number 109 to the string "109" actually requires 4 bytes: '1', '0', '9' and a terminating '\0'. Because temp[] is allocated with 3 bytes only, the final '\0' would definitely be written to temp3, which is just the m(byte#1), which unfortunately stores m's value. As a result, m's value is reset to 0 every time.
3).但是，程序员的期望是，overflow_it()中的for循环将仅执行6次，每次将m递增1.由于m始终重置为0，因此实际循环时间远远超过6倍.
3). The programmer's expectation, however, is that the for loop in the overflow_it() would execute 6 times only, with each time m being incremented by 1. Because m is always reset to 0, the actual loop time is far more than 6 times.
4).让我们看一下overflow_it()中的变量i:每次执行for循环时，i的值都会增加2，并且将访问A15Result [i].但是，如果编译并运行该程序，您将看到i值最终加起来为24，这意味着overflow_it()将数据写入A15Result [0]至A15Result [23]范围的字节中.请注意，对象str仅落后A15Result [0] 16个字节，因此overflow_it()已清除" str并破坏了其正确的内存布局.
4). Let's look at the variable i in overflow_it(): Every time the for loop is executed, i's value is incremented by 2, and A15Result[i] will be accessed. However, if you compile and run this program, you'll see the i value finally adds up to 24, which means the overflow_it() writes data to the bytes ranging from A15Result[0] to A15Result[23]. Note that the object str is only 16 bytes behind A15Result[0], thus the overflow_it() has "sweeped through" str and destroy it's correct memory layout.
5).我认为对std :: string的正确使用(因为它是非POD数据结构)取决于实例化的std :: string对象必须具有正确的内部状态.但是在此程序中，str的内部布局已在外部强制更改.这就是为什么assign()方法调用最终会导致段错误的原因.
5). I think the correct use of std::string, as it is a non-POD data structure, depends on that that instantiated std::string object must have a correct internal state. But in this program, str's internal layout has been changed by force externally. This should be why the assign() method call would finally cause a segfault.
2011年8月26日更新
在我于2011年8月19日进行的上一次更新中，我说过segfault是由对本地std :: string对象的一种方法调用引起的，该对象的内存布局已损坏，因此成为被破坏"的对象.这不是一个永远"的真实故事.考虑下面的C ++程序:
In my previous update on 08/19/2011, I said that the segfault was caused by a method call on a local std::string object whose memory layout had been broken and thus became a "destroyed" object. This is not an "always" true story. Consider the C++ program below:
//C++
class A {
public:
void Hello(const std::string& name) {
std::cout << "hello " << name;
}
};
int main(int argc, char** argv)
{
A* pa = NULL; //!!
pa->Hello("world");
return 0;
}
Hello()调用将成功.即使您为pa分配了明显错误的指针，它也会成功.原因是:根据C ++对象模型，类的非虚拟方法不在对象的内存布局中. C ++编译器将A :: Hello()方法转换为类似A_Hello_xxx(A * const this，...)的方法，该方法可以是全局函数.因此，只要您不对"this"指针进行操作，事情就可以顺利进行.
The Hello() call would succeed. It would succeed even if you assign an obviously bad pointer to pa. The reason is: the non-virtual methods of a class don't reside within the memory layout of the object, according to the C++ object model. The C++ compiler turns the A::Hello() method to something like, say, A_Hello_xxx(A * const this, ...) which could be a global function. Thus, as long as you don't operate on the "this" pointer, things could go pretty well.
此事实表明，不良"对象是否的根本原因，导致SIGSEGV segfault.在std :: string中，assign()方法不是虚拟的，因此错误的" std :: string对象不会导致segfault.可能还有其他原因最终导致了段错误.
This fact shows that a "bad" object is NOT the root cause that results in the SIGSEGV segfault. The assign() method is not virtual in std::string, thus the "bad" std::string object wouldn't cause the segfault. There must be some other reason that finally caused the segfault.
我注意到该段错误来自__gnu_cxx :: __ exchange_and_add()函数，因此我在此网页:
I noticed that the segfault comes from the __gnu_cxx::__exchange_and_add() function, so I then looked into its source code in this web page:
00046 static inline _Atomic_word
00047 __exchange_and_add(volatile _Atomic_word* __mem, int __val)
00048 { return __sync_fetch_and_add(__mem, __val); }
__ exchange_and_add()最后调用__sync_fetch_and_add().根据此网页，__sync_fetch_and_add( )是GCC内置函数，其行为如下:
The __exchange_and_add() finally calls the __sync_fetch_and_add(). According to this web page, the __sync_fetch_and_add() is a GCC builtin function whose behavior is like this:
type __sync_fetch_and_add (type *ptr, type value, ...)
{
tmp = *ptr;
*ptr op= value; // Here the "op=" means "+=" as this function is "_and_add".
return tmp;
}
有！传入的ptr指针在这里被取消引用.在08/19/2011程序中，ptr实际上是assign()方法中错误" std :: string对象的"this"指针.正是在这一点上的缺失，才真正导致了SIGSEGV分段错误.
There it is! The passed-in ptr pointer is dereferenced here. In the 08/19/2011 program, the ptr is actually the "this" pointer of the "bad" std::string object within the assign() method. It is the derefenence at this point that actually caused the SIGSEGV segmentation fault.
我们可以使用以下程序对此进行测试:
We could test this with the following program:
#include <bits/atomicity.h>
int main(int argc, char * argv[])
{
__sync_fetch_and_add((_Atomic_word *)0, 10); // Would result in a segfault.
return 0;
}
推荐答案
有两种可能:
- 第798行之前的某些代码损坏了本地
tmpTimeStamp对象 -
FormatTimeStamp()的返回值不正确.
- some code before line 798 has corrupted the local
tmpTimeStampobject - the return value from
FormatTimeStamp()was somehow bad.
_GLIBCXX_FULLY_DYNAMIC_STRING很可能是红色鲱鱼，与问题无关.
The _GLIBCXX_FULLY_DYNAMIC_STRING is most likely a red herring and has nothing to do with the problem.
如果您为libstdc++安装了debuginfo软件包(我不知道CentOS上的名称)，则可以查看"该代码，并且可以判断是否左侧的分配操作员的手侧(LHS)或RHS引起了问题.
If you install debuginfo package for libstdc++ (I don't know what it's called on CentOS), you'll be able to "see into" that code, and might be able to tell whether the left-hand-side (LHS) or the RHS of the assignment operator caused the problem.
如果无法实现，则必须在程序集级别进行调试.进入帧#2并执行x/4x $ebp应该会给您先前的ebp，呼叫者地址(0x081402fc)，LHS(应与帧#3中的&tmpTimeStamp相匹配)和RHS.从那里走，祝你好运！
If that's not possible, you'll have to debug this at the assembly level. Going into frame #2 and doing x/4x $ebp should give you previous ebp, caller address (0x081402fc), LHS (should match &tmpTimeStamp in frame #3), and RHS. Go from there, and good luck!
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